Given,
T=500 K
P1=5atm
P2=1 atm
a) For reversible isothermal process:
              Â
w= -RT ln (P2/P1)
              Â
=-8.314*500*ln(1/5)=6690.4 J
              Â
q=-w=-6690.4 J
              Â
dU=0
              Â
dH=dU-w=0+6690.4=-6690.4 J
              Â
dT= 0 for isothermal process
b) For reversible adiabatic process:
              Â
w=-RT1((P2/P1Â)(n-1)/n
-1)
              Â
=-8.314*500*((1/5)0.67/1.67-1)
              Â
=-1973 J
              Â
q=0 for adiabatic process
              Â
dU=-w=1973 J
              Â
dH=0
              Â
dT= w/CV
                      Â
=1973/20.8=94.8 K